document.write( "Question 688869: How do i find the Center, Vertices, Foci and the Equation of the asymptotes of the following equation
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document.write( "(((x-2)^2)/25)-(((y-1)^2)/9) = 1 \n" );
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Algebra.Com's Answer #425715 by lwsshak3(11628)![]() ![]() ![]() You can put this solution on YOUR website! How do i find the Center, Vertices, Foci and the Equation of the asymptotes of the following equation \n" ); document.write( "(((x-2)^2)/25)-(((y-1)^2)/9) = 1 \n" ); document.write( "This is an equation of a hyperbola with horizontal transverse axis. (x-term listed first) \n" ); document.write( "Its standard form of equation: \n" ); document.write( "For given equation: (x-2)^2)/25)-(y-1)^2)/9)=1 \n" ); document.write( "center: (2,1) \n" ); document.write( "a^2=25 \n" ); document.write( "a=√25=5 \n" ); document.write( "vertices:(2±a,1)=(2±5,1)=(-3,1) and (7,1) \n" ); document.write( "b^2=9 \n" ); document.write( "b=√9=3 \n" ); document.write( "c^2=a^2+b^2=25+9=34 \n" ); document.write( "c=√34≈5.8 \n" ); document.write( "foci:(2±c,1)=(2±5.8,1)=(-3.8,1) and (7.8,1) \n" ); document.write( "asymptotes: (straight line equations that go thru center, y=mx+b, m=slope, b=y-intercept) \n" ); document.write( "slopes of asymptotes for hyperbolas with horizontal transverse axis=±b/a=±3/5 \n" ); document.write( "Equation of asymptote with negative slope: \n" ); document.write( "y=-3x/5+b \n" ); document.write( "solve for b using coordinates of center(2,1) \n" ); document.write( "1=-3*2/5+b \n" ); document.write( "b=1+6/5=11/5 \n" ); document.write( "equation: y=-3x/5+11/5 \n" ); document.write( ".. \n" ); document.write( "Equation of asymptote with positive slope: \n" ); document.write( "y=3x/5+b \n" ); document.write( "solve for b using coordinates of center(2,1) \n" ); document.write( "1=3*2/5+b \n" ); document.write( "b=1-6/5=-1/5 \n" ); document.write( "equation: y=-3x/5-1/5 \n" ); document.write( " \n" ); document.write( " |